\(\int (d+e x^2) \sqrt {9+x^4} \, dx\) [249]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [C] (verification not implemented)
Sympy [C] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 17, antiderivative size = 163 \[ \int \left (d+e x^2\right ) \sqrt {9+x^4} \, dx=\frac {18 e x \sqrt {9+x^4}}{5 \left (3+x^2\right )}+\frac {1}{15} x \left (5 d+3 e x^2\right ) \sqrt {9+x^4}-\frac {18 \sqrt {3} e \left (3+x^2\right ) \sqrt {\frac {9+x^4}{\left (3+x^2\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt {3}}\right )|\frac {1}{2}\right )}{5 \sqrt {9+x^4}}+\frac {\sqrt {3} (5 d+9 e) \left (3+x^2\right ) \sqrt {\frac {9+x^4}{\left (3+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt {3}}\right ),\frac {1}{2}\right )}{5 \sqrt {9+x^4}} \] Output:

18*e*x*(x^4+9)^(1/2)/(5*x^2+15)+1/15*x*(3*e*x^2+5*d)*(x^4+9)^(1/2)-18/5*3^ 
(1/2)*e*(x^2+3)*((x^4+9)/(x^2+3)^2)^(1/2)*EllipticE(sin(2*arctan(1/3*x*3^( 
1/2))),1/2*2^(1/2))/(x^4+9)^(1/2)+1/5*3^(1/2)*(5*d+9*e)*(x^2+3)*((x^4+9)/( 
x^2+3)^2)^(1/2)*InverseJacobiAM(2*arctan(1/3*x*3^(1/2)),1/2*2^(1/2))/(x^4+ 
9)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 5.61 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.27 \[ \int \left (d+e x^2\right ) \sqrt {9+x^4} \, dx=3 d x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {x^4}{9}\right )+e x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {x^4}{9}\right ) \] Input:

Integrate[(d + e*x^2)*Sqrt[9 + x^4],x]
 

Output:

3*d*x*Hypergeometric2F1[-1/2, 1/4, 5/4, -1/9*x^4] + e*x^3*Hypergeometric2F 
1[-1/2, 3/4, 7/4, -1/9*x^4]
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1491, 27, 1512, 27, 761, 1510}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt {x^4+9} \left (d+e x^2\right ) \, dx\)

\(\Big \downarrow \) 1491

\(\displaystyle \frac {1}{15} \int \frac {18 \left (3 e x^2+5 d\right )}{\sqrt {x^4+9}}dx+\frac {1}{15} x \sqrt {x^4+9} \left (5 d+3 e x^2\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {6}{5} \int \frac {3 e x^2+5 d}{\sqrt {x^4+9}}dx+\frac {1}{15} x \sqrt {x^4+9} \left (5 d+3 e x^2\right )\)

\(\Big \downarrow \) 1512

\(\displaystyle \frac {6}{5} \left ((5 d+9 e) \int \frac {1}{\sqrt {x^4+9}}dx-9 e \int \frac {3-x^2}{3 \sqrt {x^4+9}}dx\right )+\frac {1}{15} x \sqrt {x^4+9} \left (5 d+3 e x^2\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {6}{5} \left ((5 d+9 e) \int \frac {1}{\sqrt {x^4+9}}dx-3 e \int \frac {3-x^2}{\sqrt {x^4+9}}dx\right )+\frac {1}{15} x \sqrt {x^4+9} \left (5 d+3 e x^2\right )\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {6}{5} \left (\frac {\left (x^2+3\right ) \sqrt {\frac {x^4+9}{\left (x^2+3\right )^2}} (5 d+9 e) \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt {3}}\right ),\frac {1}{2}\right )}{2 \sqrt {3} \sqrt {x^4+9}}-3 e \int \frac {3-x^2}{\sqrt {x^4+9}}dx\right )+\frac {1}{15} x \sqrt {x^4+9} \left (5 d+3 e x^2\right )\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {6}{5} \left (\frac {\left (x^2+3\right ) \sqrt {\frac {x^4+9}{\left (x^2+3\right )^2}} (5 d+9 e) \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt {3}}\right ),\frac {1}{2}\right )}{2 \sqrt {3} \sqrt {x^4+9}}-3 e \left (\frac {\sqrt {3} \left (x^2+3\right ) \sqrt {\frac {x^4+9}{\left (x^2+3\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt {3}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+9}}-\frac {x \sqrt {x^4+9}}{x^2+3}\right )\right )+\frac {1}{15} x \sqrt {x^4+9} \left (5 d+3 e x^2\right )\)

Input:

Int[(d + e*x^2)*Sqrt[9 + x^4],x]
 

Output:

(x*(5*d + 3*e*x^2)*Sqrt[9 + x^4])/15 + (6*(-3*e*(-((x*Sqrt[9 + x^4])/(3 + 
x^2)) + (Sqrt[3]*(3 + x^2)*Sqrt[(9 + x^4)/(3 + x^2)^2]*EllipticE[2*ArcTan[ 
x/Sqrt[3]], 1/2])/Sqrt[9 + x^4]) + ((5*d + 9*e)*(3 + x^2)*Sqrt[(9 + x^4)/( 
3 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[3]], 1/2])/(2*Sqrt[3]*Sqrt[9 + x^4]) 
))/5
 

Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 761
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1491
Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[x*( 
d*(4*p + 3) + e*(4*p + 1)*x^2)*((a + c*x^4)^p/((4*p + 1)*(4*p + 3))), x] + 
Simp[2*(p/((4*p + 1)*(4*p + 3)))   Int[Simp[2*a*d*(4*p + 3) + (2*a*e*(4*p + 
 1))*x^2, x]*(a + c*x^4)^(p - 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c 
*d^2 + a*e^2, 0] && GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]
 

rule 1510
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

rule 1512
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + c*x^4], x], x] - Simp[e/q 
 Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, c 
, d, e}, x] && PosQ[c/a]
 
Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.20

method result size
meijerg \(e \,x^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -\frac {x^{4}}{9}\right )+3 d x \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], -\frac {x^{4}}{9}\right )\) \(33\)
risch \(\frac {x \left (3 e \,x^{2}+5 d \right ) \sqrt {x^{4}+9}}{15}+\frac {2 d \sqrt {-3 i x^{2}+9}\, \sqrt {3 i x^{2}+9}\, \operatorname {EllipticF}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )}{3 \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ) \sqrt {x^{4}+9}}+\frac {6 i e \sqrt {-3 i x^{2}+9}\, \sqrt {3 i x^{2}+9}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )\right )}{5 \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ) \sqrt {x^{4}+9}}\) \(165\)
elliptic \(\frac {e \,x^{3} \sqrt {x^{4}+9}}{5}+\frac {d x \sqrt {x^{4}+9}}{3}+\frac {2 d \sqrt {-3 i x^{2}+9}\, \sqrt {3 i x^{2}+9}\, \operatorname {EllipticF}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )}{3 \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ) \sqrt {x^{4}+9}}+\frac {6 i e \sqrt {-3 i x^{2}+9}\, \sqrt {3 i x^{2}+9}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )\right )}{5 \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ) \sqrt {x^{4}+9}}\) \(169\)
default \(d \left (\frac {x \sqrt {x^{4}+9}}{3}+\frac {2 \sqrt {-3 i x^{2}+9}\, \sqrt {3 i x^{2}+9}\, \operatorname {EllipticF}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )}{3 \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ) \sqrt {x^{4}+9}}\right )+e \left (\frac {x^{3} \sqrt {x^{4}+9}}{5}+\frac {6 i \sqrt {-3 i x^{2}+9}\, \sqrt {3 i x^{2}+9}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ), i\right )\right )}{5 \left (\frac {\sqrt {6}}{6}+\frac {i \sqrt {6}}{6}\right ) \sqrt {x^{4}+9}}\right )\) \(171\)

Input:

int((e*x^2+d)*(x^4+9)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

e*x^3*hypergeom([-1/2,3/4],[7/4],-1/9*x^4)+3*d*x*hypergeom([-1/2,1/4],[5/4 
],-1/9*x^4)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.43 \[ \int \left (d+e x^2\right ) \sqrt {9+x^4} \, dx=\frac {162 i \, \sqrt {3 i} e x E(\arcsin \left (\frac {\sqrt {3 i}}{x}\right )\,|\,-1) + 6 i \, \sqrt {3 i} {\left (5 \, d - 27 \, e\right )} x F(\arcsin \left (\frac {\sqrt {3 i}}{x}\right )\,|\,-1) + {\left (3 \, e x^{4} + 5 \, d x^{2} + 54 \, e\right )} \sqrt {x^{4} + 9}}{15 \, x} \] Input:

integrate((e*x^2+d)*(x^4+9)^(1/2),x, algorithm="fricas")
 

Output:

1/15*(162*I*sqrt(3*I)*e*x*elliptic_e(arcsin(sqrt(3*I)/x), -1) + 6*I*sqrt(3 
*I)*(5*d - 27*e)*x*elliptic_f(arcsin(sqrt(3*I)/x), -1) + (3*e*x^4 + 5*d*x^ 
2 + 54*e)*sqrt(x^4 + 9))/x
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.78 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.44 \[ \int \left (d+e x^2\right ) \sqrt {9+x^4} \, dx=\frac {3 d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{9}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {3 e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{9}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \] Input:

integrate((e*x**2+d)*(x**4+9)**(1/2),x)
 

Output:

3*d*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), x**4*exp_polar(I*pi)/9)/(4*gam 
ma(5/4)) + 3*e*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), x**4*exp_polar(I 
*pi)/9)/(4*gamma(7/4))
 

Maxima [F]

\[ \int \left (d+e x^2\right ) \sqrt {9+x^4} \, dx=\int { \sqrt {x^{4} + 9} {\left (e x^{2} + d\right )} \,d x } \] Input:

integrate((e*x^2+d)*(x^4+9)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(x^4 + 9)*(e*x^2 + d), x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \left (d+e x^2\right ) \sqrt {9+x^4} \, dx=\int { \sqrt {x^{4} + 9} {\left (e x^{2} + d\right )} \,d x } \] Input:

integrate((e*x^2+d)*(x^4+9)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(x^4 + 9)*(e*x^2 + d), x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (d+e x^2\right ) \sqrt {9+x^4} \, dx=\int \sqrt {x^4+9}\,\left (e\,x^2+d\right ) \,d x \] Input:

int((x^4 + 9)^(1/2)*(d + e*x^2),x)
 

Output:

int((x^4 + 9)^(1/2)*(d + e*x^2), x)
 

Reduce [F]

\[ \int \left (d+e x^2\right ) \sqrt {9+x^4} \, dx=\frac {\sqrt {x^{4}+9}\, d x}{3}+\frac {\sqrt {x^{4}+9}\, e \,x^{3}}{5}+6 \left (\int \frac {\sqrt {x^{4}+9}}{x^{4}+9}d x \right ) d +\frac {18 \left (\int \frac {\sqrt {x^{4}+9}\, x^{2}}{x^{4}+9}d x \right ) e}{5} \] Input:

int((e*x^2+d)*(x^4+9)^(1/2),x)
 

Output:

(5*sqrt(x**4 + 9)*d*x + 3*sqrt(x**4 + 9)*e*x**3 + 90*int(sqrt(x**4 + 9)/(x 
**4 + 9),x)*d + 54*int((sqrt(x**4 + 9)*x**2)/(x**4 + 9),x)*e)/15